Đáp án:

b) \(Min =  - \dfrac{1}{3}\)

Giải thích các bước giải:

\(\begin{array}{l}
a)A = \dfrac{{\sqrt x }}{{2\sqrt x  + 1}} + \dfrac{{\left( {\sqrt x  + 1} \right)\left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {2\sqrt x  + 1} \right)}}.\left[ {\dfrac{{2x\sqrt x  - x - \sqrt x }}{{\left( {\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}} - \dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)}}{{\left( {\sqrt x  - 1} \right)\left( {\sqrt x  + 1} \right)}}} \right]\\
 = \dfrac{{\sqrt x }}{{2\sqrt x  + 1}} + \dfrac{{\sqrt x  + 1}}{{2\sqrt x  + 1}}.\dfrac{{2x\sqrt x  - x - \sqrt x  - \sqrt x \left( {x - \sqrt x  + 1} \right)}}{{\left( {\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x }}{{2\sqrt x  + 1}} + \dfrac{{\sqrt x  + 1}}{{2\sqrt x  + 1}}.\dfrac{{x\sqrt x  - 2\sqrt x }}{{\left( {\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x }}{{2\sqrt x  + 1}} + \dfrac{{x\sqrt x  - 2\sqrt x }}{{\left( {2\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x \left( {x - \sqrt x  + 1} \right) + x\sqrt x  - 2\sqrt x }}{{\left( {2\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}\\
 = \dfrac{{x\sqrt x  - x + \sqrt x  + x\sqrt x  - 2\sqrt x }}{{\left( {2\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}\\
 = \dfrac{{2x\sqrt x  - x - \sqrt x }}{{\left( {2\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}\\
 = \dfrac{{\sqrt x \left( {\sqrt x  - 1} \right)\left( {2\sqrt x  + 1} \right)}}{{\left( {2\sqrt x  + 1} \right)\left( {x - \sqrt x  + 1} \right)}}\\
 = \dfrac{{x - \sqrt x }}{{x - \sqrt x  + 1}}\\
Thay:x = 7 - 4\sqrt 3 \\
 = 4 - 2.2.\sqrt 3  + 3\\
 = {\left( {2 - \sqrt 3 } \right)^2}\\
 \to A = \dfrac{{7 - 4\sqrt 3  - \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}} }}{{7 - 4\sqrt 3  - \sqrt {{{\left( {2 - \sqrt 3 } \right)}^2}}  + 1}} = \dfrac{{1 - \sqrt 3 }}{3}\\
2)A = \dfrac{{x - \sqrt x  + 1 - 1}}{{x - \sqrt x  + 1}} = 1 - \dfrac{1}{{x - \sqrt x  + 1}}\\
 = 1 - \dfrac{1}{{x - 2\sqrt x .\dfrac{1}{2} + \dfrac{1}{4} + \dfrac{3}{4}}}\\
 = 1 - \dfrac{1}{{{{\left( {\sqrt x  - \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}}\\
Do:{\left( {\sqrt x  - \dfrac{1}{2}} \right)^2} \ge 0\forall x \ge 0\\
 \to {\left( {\sqrt x  - \dfrac{1}{2}} \right)^2} + \dfrac{3}{4} \ge \dfrac{3}{4}\\
 \to \dfrac{1}{{{{\left( {\sqrt x  - \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}} \le \dfrac{4}{3}\\
 \to  - \dfrac{1}{{{{\left( {\sqrt x  - \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}} \ge  - \dfrac{4}{3}\\
 \to 1 - \dfrac{1}{{{{\left( {\sqrt x  - \dfrac{1}{2}} \right)}^2} + \dfrac{3}{4}}} \ge  - \dfrac{1}{3}\\
 \to Min =  - \dfrac{1}{3}\\
 \Leftrightarrow \sqrt x  - \dfrac{1}{2} = 0\\
 \to x = \dfrac{1}{4}
\end{array}\)